2024 IB DP Mathematics - Analysis and Approaches 模擬試題連答案詳解

To find the area of the region, we compute the definite integral of \( f(x) \) from \( x = 0 \) to \( x = \sqrt{\ln 3} \):

\( A = \int_{0}^{\sqrt{\ln 3}} x e^{-x^2} \, dx \)

We use integration by substitution.
Let \( u = -x^2 \), which gives \( \frac{du}{dx} = -2x \), or \( x \, dx = -\frac{1}{2} \, du \).

Next, we determine the new limits of integration:
- When \( x = 0 \), \( u = 0 \).
- When \( x = \sqrt{\ln 3} \), \( u = -(\sqrt{\ln 3})^2 = -\ln 3 \).

Substituting these into the integral gives:

\( A = \int_{0}^{-\ln 3} e^u \left(-\frac{1}{2}\right) \, du \)
\( A = \frac{1}{2} \int_{-\ln 3}^{0} e^u \, du \)

Now, evaluate the integral:

\( A = \frac{1}{2} \left[ e^u \right]_{-\ln 3}^{0} \)
\( A = \frac{1}{2} \left( e^0 - e^{-\ln 3} \right) \)
\( A = \frac{1}{2} \left( 1 - \frac{1}{3} \right) \)
\( A = \frac{1}{2} \left( \frac{2}{3} \right) = \frac{1}{3} \)

M1: Setting up the correct integral expression: \( \int_{0}^{\sqrt{\ln 3}} x e^{-x^2} \, dx \).
M1.5: Attempting substitution, identifying \( u = -x^2 \) (or \( u = x^2 \)) and obtaining \( x \, dx = -\frac{1}{2} \, du \) (or equivalent).
M1: Correctly changing the integration limits to \( 0 \) and \( -\ln 3 \) (or converting back to \( x \) after integrating).
A1: Finding the antiderivative \( -\frac{1}{2} e^{-x^2} \) (or \( -\frac{1}{2} e^u \)).
A1: Substituting limits correctly to get \( \frac{1}{3} \) (or equivalent).

(a) To find the range, we examine the behavior of \( f(x) \) as \( x \to \infty \) and \( x \to -\infty \).

As \( x \to \infty \), \( e^x \to \infty \), so we divide numerator and denominator by \( e^x \):
\( \lim_{x \to \infty} \frac{3 - e^{-x}}{1 + 2e^{-x}} = \frac{3 - 0}{1 + 0} = 3 \).

As \( x \to -\infty \), \( e^x \to 0 \):
\( \lim_{x \to -\infty} \frac{3e^x - 1}{e^x + 2} = \frac{0 - 1}{0 + 2} = -\frac{1}{2} \).

Since \( f(x) \) is continuous and strictly increasing (as the derivative is positive for all \( x \)), the range of \( f \) is \( -\frac{1}{2} < y < 3 \).

(b) To find the expression for the inverse function, let \( y = f(x) \) and swap \( x \) and \( y \):

\( x = \frac{3e^y - 1}{e^y + 2} \)

Multiply both sides by \( e^y + 2 \):

\( x(e^y + 2) = 3e^y - 1 \)
\( x e^y + 2x = 3e^y - 1 \)

Rearrange to group terms with \( e^y \) on one side:

\( 2x + 1 = 3e^y - x e^y \)
\( 2x + 1 = e^y (3 - x) \)

Isolate \( e^y \):

\( e^y = \frac{2x + 1}{3 - x} \)

Take the natural logarithm of both sides:

\( y = \ln\left(\frac{2x + 1}{3 - x}\right) \)

Thus, \( f^{-1}(x) = \ln\left(\frac{2x+1}{3-x}\right) \).

(a)
M1: Evaluating the limit of \( f(x) \) as \( x \to \infty \) to find the upper bound 3.
M1: Evaluating the limit of \( f(x) \) as \( x \to -\infty \) to find the lower bound \( -\frac{1}{2} \).
R0.5: Correctly reasoning or showing why the values lie strictly between these limits (e.g., stating \( e^x > 0 \)).

(b)
M1: Setting up the equation by swapping variables: \( x = \frac{3e^y - 1}{e^y + 2} \).
M1: Rearranging the equation to group \( e^y \) terms together: \( e^y(3 - x) = 2x + 1 \).
A1: Correctly writing the final expression for \( f^{-1}(x) \): \( f^{-1}(x) = \ln\left(\frac{2x+1}{3-x}\right) \).

Since the sum of probabilities must equal 1:
\( p + q + r = 1 \)

We are given that \( P(X \leq 2) = p + q = 0.7 \).
Using this in the sum of probabilities equation:
\( 0.7 + r = 1 \implies r = 0.3 \)

The expectation of \( X \) is given by:
\( \text{E}(X) = 1(p) + 2(q) + 3(r) = 2.1 \)

Substitute \( r = 0.3 \) into the expectation formula:
\( p + 2q + 3(0.3) = 2.1 \)
\( p + 2q + 0.9 = 2.1 \)
\( p + 2q = 1.2 \)

We now have a system of linear equations:
1) \( p + q = 0.7 \)
2) \( p + 2q = 1.2 \)

Subtract the first equation from the second:
\( (p + 2q) - (p + q) = 1.2 - 0.7 \)
\( q = 0.5 \)

Substitute \( q = 0.5 \) back into the first equation:
\( p + 0.5 = 0.7 \implies p = 0.2 \)

Thus, the values are:
\( p = 0.2 \), \( q = 0.5 \), \( r = 0.3 \).

M1.5: Using \( P(X \leq 2) = 0.7 \) along with \( p+q+r=1 \) to determine \( r = 0.3 \).
M1: Correctly writing the expectation formula: \( p + 2q + 3r = 2.1 \).
M1: Substituting \( r = 0.3 \) to obtain the simplified linear equation \( p + 2q = 1.2 \).
M1: Setting up a valid system of equations to solve for \( p \) and \( q \).
A1: Successfully calculating both values: \( p = 0.2 \) and \( q = 0.5 \).

First, divide both sides of the equation by \( \cos(2x) \) (noting that \( \cos(2x) \neq 0 \) since if it were, \( \sin(2x) \) would also have to be zero, which is impossible):

\( \sqrt{3} \frac{\sin(2x)}{\cos(2x)} = 1 \)
\( \sqrt{3}\tan(2x) = 1 \)
\( \tan(2x) = \frac{1}{\sqrt{3}} \)

Since \( 0 \leq x \leq \pi \), the interval for \( 2x \) is:
\( 0 \leq 2x \leq 2\pi \)

We look for values in the first and third quadrants where the tangent function is positive and equals \( \frac{1}{\sqrt{3}} \):

\( 2x = \frac{\pi}{6} \) or \( 2x = \pi + \frac{\pi}{6} = \frac{7\pi}{6} \)

Now, divide both values by 2 to find \( x \):

\( x = \frac{\pi}{12} \) or \( x = \frac{7\pi}{12} \)

Both values satisfy the original condition \( 0 \leq x \leq \pi \).

M1.5: Dividing both sides to form a tangent equation, e.g., \( \tan(2x) = \frac{1}{\sqrt{3}} \).
M1: Identifying the correct domain for the double angle: \( 0 \leq 2x \leq 2\pi \).
M1.5: Finding the correct angles for \( 2x \): \( 2x = \frac{\pi}{6} \) and \( 2x = \frac{7\pi}{6} \) (award 0.75 marks for each correct angle).
A1.5: Dividing by 2 to find the exact values of \( x \): \( x = \frac{\pi}{12} \) and \( x = \frac{7\pi}{12} \) (award 0.75 marks for each correct final answer).

(a) For a geometric sequence, the \( n \)-th term is given by \( u_n = u_1 r^{n-1} \).
We are given:
\( u_3 = u_1 r^2 = 24 \)
\( u_6 = u_1 r^5 = 3 \)

Dividing the second equation by the first:
\( \frac{u_1 r^5}{u_1 r^2} = \frac{3}{24} \)
\( r^3 = \frac{1}{8} \)
\( r = \sqrt[3]{\frac{1}{8}} = \frac{1}{2} \)

(b) Substitute \( r = \frac{1}{2} \) into the equation for \( u_3 \):
\( u_1 \left(\frac{1}{2}\right)^2 = 24 \)
\( u_1 \left(\frac{1}{4}\right) = 24 \)
\( u_1 = 24 \times 4 = 96 \)

(c) The sum to infinity is given by:
\( S_{\infty} = \frac{u_1}{1 - r} \)

Since \( |r| < 1 \), we can compute:
\( S_{\infty} = \frac{96}{1 - 1/2} = \frac{96}{1/2} = 192 \)

(a)
M1: Expressing terms using the geometric sequence formula to form the ratio: \( \frac{u_1 r^5}{u_1 r^2} = \frac{3}{24} \).
A1: Solving for \( r \) to obtain \( r = \frac{1}{2} \).

(b)
M1: Substituting the value of \( r \) into the equation for \( u_3 \) or \( u_6 \).
A0.5: Finding \( u_1 = 96 \).

(c)
M1: Recalling and substituting into the sum to infinity formula: \( S_{\infty} = \frac{96}{1 - 1/2} \).
A1: Correctly evaluating to get \( 192 \).

First, find the \( y \)-coordinate of the point of contact by substituting \( x = e \) into \( g(x) \):
\( g(e) = e^2 \ln(e) = e^2 \cdot 1 = e^2 \).
So, the point of contact is \( (e, e^2) \).

Next, we find the derivative \( g'(x) \) using the product rule:
\( g'(x) = \frac{d}{dx}(x^2) \cdot \ln(x) + x^2 \cdot \frac{d}{dx}(\ln x) \)
\( g'(x) = 2x \ln(x) + x^2 \cdot \frac{1}{x} \)
\( g'(x) = 2x \ln(x) + x \)

Evaluate the derivative at \( x = e \) to find the gradient of the tangent, \( m \):
\( m = g'(e) = 2e \ln(e) + e = 2e(1) + e = 3e \).

The equation of the tangent line can be written using the point-gradient formula \( y - y_1 = m(x - x_1) \):
\( y - e^2 = 3e(x - e) \)

Expanding the right-hand side:
\( y - e^2 = 3ex - 3e^2 \)

Isolating \( y \) to match the form \( y = mx + c \):
\( y = 3ex - 3e^2 + e^2 \)
\( y = 3ex - 2e^2 \)

M1: Finding the \( y \)-coordinate of the point of contact, yielding \( (e, e^2) \).
M1.5: Applying the product rule to find \( g'(x) = 2x \ln(x) + x \) (award 0.75 marks for each part of the sum).
M1: Evaluating the derivative at \( x = e \) to find the gradient \( m = 3e \).
M1: Substituting the point and gradient into a linear equation format, e.g., \( y - e^2 = 3e(x - e) \).
A1: Simplifying into the required format: \( y = 3ex - 2e^2 \).

(a) To find the \(x\)-intercepts, we set \(f(x) = 0\):
\(e^{2x} - 4e^x + 3 = 0\)
Let \(u = e^x\):
\(u^2 - 4u + 3 = 0 \Rightarrow (u-1)(u-3) = 0\)
This gives \(e^x = 1\) or \(e^x = 3\).
Therefore, \(x = 0\) or \(x = \ln 3\).
So the \(x\)-intercepts are at \((0,0)\) and \((\ln 3, 0)\).

(b) First, we find the derivative \(f'(x)\):
\(f'(x) = 2e^{2x} - 4e^x\)
To find the critical points, set \(f'(x) = 0\):
\(2e^x(e^x - 2) = 0\)
Since \(e^x > 0\) for all real \(x\), we must have \(e^x = 2 \Rightarrow x = \ln 2\).
Now we find the corresponding \(y\)-coordinate:
\(f(\ln 2) = e^{2\ln 2} - 4e^{\ln 2} + 3 = 4 - 4(2) + 3 = -1\).
To justify that \((\ln 2, -1)\) is a local minimum, we can use the second derivative:
\(f''(x) = 4e^{2x} - 4e^x\)
\(f''(\ln 2) = 4(4) - 4(2) = 16 - 8 = 8 > 0\).
Since the second derivative is positive at this point, it is a local minimum.
Thus, the coordinates of the local minimum are \((\ln 2, -1)\).

(c) At \(x = 0\), the gradient of the tangent is:
\(f'(0) = 2e^0 - 4e^0 = -2\).
Therefore, the gradient of the normal to the curve is the negative reciprocal:
\(m_{\text{normal}} = -\frac{1}{f'(0)} = \frac{1}{2}\).
Since \(f(0) = 0\), the normal passes through \((0, 0)\).
The equation of the normal is:
\(y - 0 = \frac{1}{2}(x - 0) \Rightarrow y = \frac{1}{2}x\).

(d) The region is completely enclosed between the \(x\)-intercepts \(x = 0\) and \(x = \ln 3\).
Since \(f(\ln 2) = -1 < 0\), the curve lies below the \(x\)-axis in this interval.
The area is given by:
\(A = \int_{0}^{\ln 3} (0 - f(x)) \mathrm{d}x = \int_{0}^{\ln 3} (4e^x - e^{2x} - 3) \mathrm{d}x\)
Integrating term by term:
\(\int (4e^x - e^{2x} - 3) \mathrm{d}x = \left[ 4e^x - \frac{1}{2}e^{2x} - 3x \right]_{0}^{\ln 3}\)
Substituting the upper limit \(x = \ln 3\):
\(4(3) - \frac{1}{2}(9) - 3\ln 3 = 12 - 4.5 - 3\ln 3 = \frac{15}{2} - 3\ln 3\)
Substituting the lower limit \(x = 0\):
\(4(1) - \frac{1}{2}(1) - 0 = \frac{7}{2}\)
Subtracting the lower limit value from the upper limit value:
\(A = \left(\frac{15}{2} - 3\ln 3\right) - \frac{7}{2} = 4 - 3\ln 3\).

(a)
M1 for attempting to solve the quadratic equation in \(e^x\) (e.g., substitution or factorizing).
A1 for \(e^x = 1\) and \(e^x = 3\).
A1 for \(x = 0\) and \(x = \ln 3\) (or coordinates \((0,0)\) and \((\ln 3, 0)\)).

(b)
M1 for finding a correct expression for \(f'(x)\).
A1 for setting \(f'(x) = 0\) and finding \(x = \ln 2\).
A1 for finding the correct \(y\)-coordinate \(y = -1\).
R1 for a valid justification (e.g., second derivative test or sign diagram showing derivative changes from negative to positive) that it is a minimum.

(c)
M1 for finding the gradient of the tangent at \(x = 0\), which is \(-2\).
M1 for finding the gradient of the normal as the negative reciprocal, \(\frac{1}{2}\).
A1 for the correct equation \(y = \frac{1}{2}x\) (accept equivalent forms).

(d)
M1 for setting up the correct integral with correct limits, \(\int_0^{\ln 3} -f(x) \mathrm{d}x\).
M1 for integrating correctly: \(4e^x - \frac{1}{2}e^{2x} - 3x\).
A1 for substituting \(x = \ln 3\) correctly to get \(\frac{15}{2} - 3\ln 3\).
A1 for substituting \(x = 0\) correctly to get \(\frac{7}{2}\).
A1 for the final simplified answer \(4 - 3\ln 3\).

(a) We start with the Left-Hand Side (LHS):
\(\text{LHS} = \cos(2\theta) + \sin(2\theta) + 1\)
Using the double-angle identities:
\(\cos(2\theta) = 2\cos^2\theta - 1\)
\(\sin(2\theta) = 2\sin\theta\cos\theta\)
Substituting these into the equation:
\(\text{LHS} = (2\cos^2\theta - 1) + 2\sin\theta\cos\theta + 1\)
\(\text{LHS} = 2\cos^2\theta + 2\sin\theta\cos\theta\)
Factoring out \(2\cos\theta\):
\(\text{LHS} = 2\cos\theta(\cos\theta + \sin\theta) = \text{RHS}\).
Thus, the identity is shown.

(b) Using the identity from part (a), the equation becomes:
\(2\cos\theta(\cos\theta + \sin\theta) = 0\)
This gives two cases to solve:

Case 1: \(\cos\theta = 0\)
For \(0 \le \theta \le \pi\), the solution is:
\(\theta = \frac{\pi}{2}\).

Case 2: \(\cos\theta + \sin\theta = 0\)
\(\sin\theta = -\cos\theta\)
\(\tan\theta = -1\)
For \(0 \le \theta \le \pi\), the solution is:
\(\theta = \frac{3\pi}{4}\).

Thus, the solutions in the given domain are \(\theta = \frac{\pi}{2}\) and \(\theta = \frac{3\pi}{4}\).

(c) (i) We want to write \(\cos(2x) + \sin(2x)\) in the form \(p\cos(2x - q)\).
Expanding \(p\cos(2x - q) = p\cos(2x)\cos q + p\sin(2x)\sin q\).
Comparing coefficients with \(1\cos(2x) + 1\sin(2x)\):
\(p\cos q = 1\)
\(p\sin q = 1\)
Squaring and adding both equations:
\(p^2(\cos^2 q + \sin^2 q) = 1^2 + 1^2 = 2 \Rightarrow p = \sqrt{2}\) (since \(p > 0\)).
Dividing the equations:
\(\frac{p\sin q}{p\cos q} = \frac{1}{1} \Rightarrow \tan q = 1\).
Since \(0 < q < \frac{\pi}{2}\), we have \(q = \frac{\pi}{4}\).
Therefore, \(g(x) = \sqrt{2}\cos\left(2x - \frac{\pi}{4}\right) + 1\).

(ii) The maximum value of the cosine function is 1.
Thus, the maximum value of \(g(x)\) is:
\(g_{\text{max}} = \sqrt{2}(1) + 1 = \sqrt{2} + 1\).
This maximum occurs when:
\(\cos\left(2x - \frac{\pi}{4}\right) = 1\)
\(2x - \frac{\pi}{4} = 0 \Rightarrow 2x = \frac{\pi}{4} \Rightarrow x = \frac{\pi}{8}\).
Since \(\frac{\pi}{8} > 0\), this is the smallest positive value of \(x\).

(a)
M1 for substituting \(\cos(2\theta) = 2\cos^2\theta - 1\) (or equivalent double-angle identity).
M1 for substituting \(\sin(2\theta) = 2\sin\theta\cos\theta\).
A1 for factoring out \(2\cos\theta\) to obtain the required form.

(b)
M1 for setting both factors to zero: \(\cos\theta = 0\) and \(\cos\theta + \sin\theta = 0\).
A1 for solving \(\cos\theta = 0\) to obtain \(\theta = \frac{\pi}{2}\).
M1 for rewriting \(\cos\theta + \sin\theta = 0\) as \(\tan\theta = -1\) (or equivalent).
A1 for solving \(\tan\theta = -1\) to obtain \(\theta = \frac{3\pi}{4}\).
A1 for stating exactly the two solutions: \(\theta = \frac{\pi}{2}\) and \(\theta = \frac{3\pi}{4}\) (and no others in the domain).

(c)(i)
M1 for equating coefficients or setting up a right-angled triangle to find \(p\) and \(q\).
A1 for finding \(p = −\sqrt{2}\).
M1 for setting up \(\tan q = 1\).
A1 for finding \(q = \frac{\pi}{4}\).

(c)(ii)
A1 for maximum value \(\sqrt{2} + 1\).
M1 for setting \(2x - \frac{\pi}{4} = 0\) (or equivalent).
A1 for finding \(x = \frac{\pi}{8}\).

(a) Let \(A\) be the event that Bag A is selected, \(B\) be the event that Bag B is selected, and \(R\) be the event that a red ball is drawn.
The probability of selecting Bag A is:
\(P(A) = \frac{2}{6} = \frac{1}{3}\)
The probability of selecting Bag B is:
\(P(B) = \frac{4}{6} = \frac{2}{3}\)

The probability of drawing a red ball from Bag A is:
\(P(R|A) = \frac{3}{5}\)
The probability of drawing a red ball from Bag B is:
\(P(R|B) = \frac{3}{6} = \frac{1}{2}\)

Using the law of total probability:
\(P(R) = P(A)P(R|A) + P(B)P(R|B)\)
\(P(R) = \left(\frac{1}{3}\right)\left(\frac{3}{5}\right) + \left(\frac{2}{3}\right)\left(\frac{1}{2}\right)\)
\(P(R) = \frac{1}{5} + \frac{1}{3} = \frac{3 + 5}{15} = \frac{8}{15}\).
Thus, the probability of drawing a red ball is \(\frac{8}{15}\).

(b) We want to find the conditional probability \(P(A|R)\):
\(P(A|R) = \frac{P(A \cap R)}{P(R)}\)
\(P(A \cap R) = P(A)P(R|A) = \frac{1}{3} \times \frac{3}{5} = \frac{1}{5}\)
\(P(A|R) = \frac{1/5}{8/15} = \frac{1}{5} \times \frac{15}{8} = \frac{3}{8}\).

(c) Since each trial is independent with a constant probability of success \(P(R) = \frac{8}{15}\), let \(X\) be the number of red balls drawn in 3 trials. Then \(X \sim \text{B}\left(3, \frac{8}{15}\right)\).
We want to find \(P(X = 2)\):
\(P(X = 2) = \binom{3}{2} \left(\frac{8}{15}\right)^2 \left(1 - \frac{8}{15}\right)\)
\(P(X = 2) = 3 \times \left(\frac{64}{225}\right) \times \left(\frac{7}{15}\right)\)
\(P(X = 2) = 3 \times \frac{448}{3375} = \frac{448}{1125}\).

(d) Let \(W\) be the winnings of a player in a single turn.
The possible outcomes and their probabilities are:
1. Bag A and Red: Player wins $5.
\(P(W = 5) = P(A \cap R) = \frac{1}{5} = \frac{3}{15}\).
2. Bag B and Red: Player wins $2.
\(P(W = 2) = P(B \cap R) = \frac{2}{3} \times \frac{1}{2} = \frac{1}{3} = \frac{5}{15}\).
3. Any blue ball: Player loses $3 (wins \-$3).
The probability of drawing a blue ball is \(1 - P(R) = 1 - \frac{8}{15} = \frac{7}{15}\).
\(P(W = -3) = \frac{7}{15}\).

To check that probabilities sum to 1:
\(\frac{3}{15} + \frac{5}{15} + \frac{7}{15} = 1\).

The expected winnings are:
\(E(W) = 5\left(\frac{3}{15}\right) + 2\left(\frac{5}{15}\right) - 3\left(\frac{7}{15}\right)\)
\(E(W) = \frac{15 + 10 - 21}{15} = \frac{4}{15}\) dollars (approximately $0.27).

(a)
M1 for determining the probability of selecting each bag: \(P(A) = 1/3\), \(P(B) = 2/3\).
M1 for determining the conditional probabilities of drawing a red ball from each bag: \(P(R|A) = 3/5\), \(P(R|B) = 1/2\).
M1 for substituting these values into the total probability formula.
A1 for showing correct intermediate steps leading to \(8/15\).

(b)
M1 for writing the correct conditional probability expression: \(P(A|R) = \frac{P(A \cap R)}{P(R)}\).
A1 for identifying \(P(A \cap R) = 1/5\).
A1 for simplifying to \(3/8\).

(c)
M1 for recognizing the binomial distribution model \(X \sim \text{B}(3, 8/15)\).
M1 for using the binomial probability formula with correct parameters.
A1 for substituting values correctly: \(3 \times (8/15)^2 \times (7/15)\).
A1 for the final answer \(\frac{448}{1125}\).

(d)
M1 for calculating the probability of winning $5: \(P(W = 5) = 1/5\).
M1 for calculating the probability of winning $2: \(P(W = 2) = 1/3\) and losing $3: \(P(W = -3) = 7/15\).
M1 for setting up the expected value equation: \(5(3/15) + 2(5/15) - 3(7/15)\).
A1 for the correct expected value \(\frac{4}{15}\) dollars (accept equivalent fractions, or $0.27, do not accept negative values).

Let \(X\) be the weight of a package of coffee, where \(X \sim N(250, 4^2)\).

(a) Using a graphic display calculator:
\(P(245 < X < 253) \approx 0.668713...\)
To 3 significant figures, the probability is 0.669.

(b) We are given that \(P(X < w) = 0.15\).
Using the inverse normal function on a graphic display calculator:
\(w \approx 245.854...\)
To 3 significant figures, \(w = 246\) g.

(a) [2 marks]
Award (M1) for writing a correct probability statement, e.g., \(P(245 < X < 253)\).
Award (A1) for 0.669 (accept 0.6687).

(b) [3.83 marks]
Award (M1) for setting up a correct equation, e.g., \(P(X < w) = 0.15\) or representing the inverse normal distribution calculation.
Award (A1) for 245.85.
Award (A1) for 246 (to 3 s.f.).

(a) The particle is at rest when \(v(t) = 0\).
\(6 \ln(t+1) - 2t = 0\)
Using a GDC to find the non-zero root:
\(t \approx 5.7448...\)
To 3 significant figures, the non-zero time when the particle is at rest is \(t = 5.74\) seconds.

(b) To find the maximum velocity, we solve \(v'(t) = 0\).
Using the chain rule to differentiate:
\(v'(t) = \frac{6}{t+1} - 2\)
Setting the derivative to zero:
\(\frac{6}{t+1} - 2 = 0 \implies t+1 = 3 \implies t = 2\)
Substitute \(t = 2\) back into the velocity function:
\(v(2) = 6 \ln(3) - 4 \approx 2.5917...\)
To 3 significant figures, the maximum velocity of the particle is 2.59 \(\text{m}\,\text{s}^{-1}\).

(a) [2.83 marks]
Award (M1) for setting \(v(t) = 0\).
Award (A1) for \(t \approx 5.74\).

(b) [3 marks]
Award (M1) for attempting to differentiate \(v(t)\) or identifying the maximum occurs when \(v'(t) = 0\).
Award (A1) for \(t = 2\).
Award (A1) for \(2.59\) (accept 2.5917).

(a) At \(t = 0\), the temperature is \(85^\circ\text{C}\).
\(85 = A e^{0} + 20\)
\(85 = A + 20 \implies A = 65\).

(b) At \(t = 5\), the temperature is \(55^\circ\text{C}\).
\(55 = 65 e^{-5k} + 20\)
\(35 = 65 e^{-5k}\)
\(e^{-5k} = \frac{35}{65} = \frac{7}{13}\)
Taking natural logarithms of both sides:
\(-5k = \ln\left(\frac{7}{13}\right)\)
\(k = -\frac{1}{5}\ln\left(\frac{7}{13}\right) \approx 0.123819...\)
To 3 significant figures, \(k = 0.124\).

(c) We find \(t\) when \(T(t) = 35\):
\(35 = 65 e^{-kt} + 20\)
\(15 = 65 e^{-kt}\)
\(e^{-kt} = \frac{15}{65} = \frac{3}{13}\)
Taking natural logarithms:
\(-kt = \ln\left(\frac{3}{13}\right)\)
\(t = -\frac{1}{0.123819...}\ln\left(\frac{3}{13}\right) \approx 11.84...\)
To 3 significant figures, \(t = 11.8\) minutes.

(a) [1.83 marks]
Award (M1) for substituting \(t=0\) and \(T=85\) into the function.
Award (A1) for \(A = 65\).

(b) [2 marks]
Award (M1) for setting up the equation \(55 = 65 e^{-5k} + 20\).
Award (A1) for \(k \approx 0.124\) (accept 0.1238).

(c) [2 marks]
Award (M1) for setting up the equation \(35 = 65 e^{-kt} + 20\).
Award (A1) for \(11.8\) (accept 11.84).

(a) Using the area formula for a triangle, \(\text{Area} = \frac{1}{2} a c \sin B\):
\(28 = \frac{1}{2} (10)(7) \sin B\)
\(28 = 35 \sin B \implies \sin B = 0.8\)
Using a GDC to find the acute angle:
\(B_1 = \arcsin(0.8) \approx 0.92729...\) radians.
The obtuse angle is:
\(B_2 = \pi - 0.92729... \approx 2.21429...\) radians.
To 3 significant figures, the possible angles are \(0.927\) and \(2.21\) radians.

(b) Since angle \(B\) is obtuse, we use \(B \approx 2.21429...\) radians (or \(\cos B = -0.6\)).
Using the cosine rule to find \(AC\):
\(AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos B\)
\(AC^2 = 7^2 + 10^2 - 2(7)(10)\cos(2.21429...)\)
\(AC^2 = 49 + 100 - 140(-0.6) = 149 + 84 = 233\)
\(AC = \sqrt{233} \approx 15.264...\)
To 3 significant figures, \(AC = 15.3\) cm.

(a) [3 marks]
Award (M1) for substituting the given area and lengths into the area formula.
Award (A1) for the acute angle \(0.927\) (accept 0.9273).
Award (A1) for the obtuse angle \(2.21\) (accept 2.214).

(b) [2.83 marks]
Award (M1) for selecting the obtuse angle or evaluating \(\cos B = -0.6\).
Award (M1) for substituting correctly into the cosine rule.
Award (A1) for \(15.3\) (accept 15.26).

(a) We use the compound interest formula: \(FV = PV \left(1 + \frac{r}{100k}\right)^{kn}\),
where \(PV = 5000\), \(r = 4.2\), \(k = 12\) (since interest is compounded monthly), and \(n = 6\).
\(FV = 5000 \left(1 + \frac{4.2}{1200}\right)^{72} = 5000(1.0035)^{72} \approx 6431.87...\)
To the nearest dollar, the amount in the account is \(\$6432\).

(b) To double in value, the future value must be at least \(\$10000\).
\(5000(1.0035)^{12y} \ge 10000 \implies (1.0035)^{12y} \ge 2\)
Taking logarithms on both sides:
\(12y \ln(1.0035) \ge \ln(2)\)
\(y \ge \frac{\ln(2)}{12 \ln(1.0035)} \approx 16.531...\)
Since we require the minimum number of complete years, we round up to the next integer:
\(y = 17\) years.

(a) [2.83 marks]
Award (M1) for correct substitution into the compound interest formula.
Award (A1) for a calculation yielding \(6431.87\).
Award (A1) for rounding to the nearest dollar: \(6432\).

(b) [3 marks]
Award (M1) for setting up the equation or inequality, e.g., \(5000(1.0035)^{12y} \ge 10000\).
Award (A1) for finding the decimal boundary \(16.5\) (or equivalent boundary in months, \(198.38\)).
Award (A1) for \(17\) years.

(a) Entering the data into a graphic display calculator:
\(r \approx 0.988889...\)
To 3 significant figures, \(r = 0.989\).

(b) The equation of the regression line is \(y = ax + b\).
From the calculator, we obtain the coefficients:
\(a \approx 3.09219...\)
\(b \approx -1.2437...\)
To 3 significant figures, the equation of the regression line is:
\(y = 3.09x - 1.24\).

(c) Substituting \(x = 22\) into the regression equation (using unrounded coefficients for precision):
\(y = 3.09219(22) - 1.2437 \approx 66.784...\)
To the nearest whole number, the estimated number of cold drinks sold is 67.

(a) [1.83 marks]
Award (A1) for 0.989 (accept 0.99).

(b) [2 marks]
Award (M1) for an attempt to find a linear regression line using GDC.
Award (A1) for \(y = 3.09x - 1.24\) (or equivalent accuracy).

(c) [2 marks]
Award (M1) for substituting \(x = 22\) into their regression equation.
Award (A1) for 67 (accept 66.8).

Let \(X\) be the mass of an apple, where \(X \sim N(150, 18^2)\). (a) Using a GDC, \(P(140 < X < 165) \approx 0.508\). (b) We are given \(P(X > M) = 0.15\), which means \(P(X \le M) = 0.85\). Using the inverse normal function on a GDC, we find \(M \approx 169\) grams. (c) Let \(Y\) be the number of Premium apples in a box of 20, where \(Y \sim B(20, 0.15)\). (i) Using binomial probability, \(P(Y = 4) = \binom{20}{4} (0.15)^4 (0.85)^{16} \approx 0.182\). (ii) We want \(P(Y \ge 2) = 1 - P(Y \le 1)\). Using binomial CDF on a GDC, \(P(Y \le 1) \approx 0.176\), so \(P(Y \ge 2) \approx 1 - 0.1755 = 0.824\). (d) For the 3rd Premium apple to be the 10th apple selected, there must be exactly 2 Premium apples in the first 9 selections, and the 10th selection must be a Premium apple. Let \(W \sim B(9, 0.15)\). The probability is \(P(W = 2) \times 0.15 = \binom{9}{2} (0.15)^2 (0.85)^7 \times 0.15 \approx 0.2597 \times 0.15 \approx 0.0390\).

(a) M1 for setting up the normal boundary, A1 for 0.508. (b) M1 for \(P(X \le M) = 0.85\), A1 for setting up inverse normal, A1 for 169. (c)(i) M1 for recognizing binomial distribution \(Y \sim B(20, 0.15)\), A1 for 0.182. (c)(ii) M1 for setting up \(1 - P(Y \le 1)\) or equivalent sum, A1 for 0.824. (d) M1 for recognizing the need for 2 Premium apples in the first 9 trials, A1 for \(P(W = 2) \approx 0.2597\), M1 for multiplying by 0.15, A1 for the final answer 0.0390.

(a) The initial velocity is \(v(0) = 3(0) - e^0 \cos(0) = -1 \text{ m s}^{-1}\). (b) The particle is at rest when \(v(t) = 0\). Using a GDC to solve \(3t - e^{0.4t}\cos(t) = 0\) in the interval \(0 \le t \le 10\), we find \(t \approx 0.360\) seconds. (c) The acceleration is the derivative of velocity, \(a(t) = v'(t)\). Using a GDC to find the numerical derivative at \(t = 4\), we get \(a(4) \approx 0.547 \text{ m s}^{-2}\). (d) The total distance travelled is given by \(\int_{0}^{10} |v(t)| dt\). Using numerical integration on a GDC, we find the total distance is approximately 192 meters. (e) To find the maximum velocity, we evaluate \(v(t)\) at critical points where \(v'(t) = 0\) and at the boundaries. The critical points of \(v(t)\) in \(0 \le t \le 10\) occur when \(v'(t) = 0\), which gives \(t \approx 4.10\) (where \(v \approx 15.3\)) and \(t \approx 9.86\) (where \(v \approx 76.4\)). At the boundaries, \(v(0) = -1\) and \(v(10) \approx 75.8\). Comparing these values, the maximum velocity is approximately 76.4 meters per second, which occurs at \(t \approx 9.86\) seconds.

(a) A1 for -1. (b) M1 for setting \(v(t) = 0\), A2 for \(t \approx 0.360\). (c) M1 for recognizing \(a(t) = v'(t)\), M1 for evaluating at \(t=4\), A1 for 0.547. (d) M1 for setting up the integral of the absolute value, A2 for splitting the integral or using GDC directly, A1 for 192. (e) M1 for finding critical points where \(v'(t) = 0\), A1 for \(t \approx 9.86\), M1 for comparing critical values and boundary values, A1 for maximum velocity of 76.4 at \(t = 9.86\).

(a)(i) The amplitude \(a = \frac{6.2 - 1.8}{2} = 2.2\). (ii) The vertical shift \(d = \frac{6.2 + 1.8}{2} = 4\). (iii) The time between a maximum and the next minimum is half the period, so \(\frac{T}{2} = 9 - 3 = 6 \implies T = 12\) hours. Thus \(b = \frac{2\pi}{12} = \frac{\rm \pi}{6} \approx 0.524\). (iv) The first maximum occurs at \(t = 3\), so \(c = 3\). The model is \(h(t) = 2.2 \cos\left(\frac{\pi}{6}(t - 3)\right) + 4\). (b) At 14:00, \(t = 14\). Substituting this into the function: \(h(14) = 2.2 \cos\left(\frac{11\pi}{6}\right) + 4 \approx 5.91\) meters. (c) We solve \(h(t) \ge 5\) over the interval \(0 \le t \le 24\). Using a GDC to find the intersection points with the line \(y = 5\), we get \(t_1 \approx 0.899\), \(t_2 \approx 5.101\), \(t_3 \approx 12.899\), and \(t_4 \approx 17.101\). The water height is at least 5 meters during the intervals \([0.899, 5.101]\) and \([12.899, 17.101]\). The total duration is \((5.101 - 0.899) + (17.101 - 12.899) = 4.202 + 4.202 = 8.40\) hours.

(a)(i) A1 for 2.2. (ii) A1 for 4. (iii) M1 for finding period \(T = 12\), A1 for \(b = \frac{\pi}{6}\) (or 0.524). (iv) A2 for \(c = 3\). (b) M1 for substituting \(t = 14\), A1 for 5.91. (c) M1 for setting up the inequality \(h(t) \ge 5\), A1 for finding the critical times in the first cycle (\(t \approx 0.899\) and \(t \approx 5.101\)), A1 for finding the critical times in the second cycle (\(t \approx 12.899\) and \(t \approx 17.101\)), M1 for calculating the interval widths, A1 for the total time of 8.40 hours.